Elementary Particle Physics Fall 2009 Lecture 08

MW 12:50 PM – 2:10 PM, Physics D‐122 Michael Rijssenbeek email: [email protected] Skype: mrijssenbeek Course Web page: http://sbhep1.physics.sunysb.edu/~rijssenbeek/PHY557_F09.html or via departmental course page or BlackBoard

10/19/2009 1 Example: a + b → 1 + 2 in CMS

The momenta |pi| and |pf| of the initial and final state particles are fully determined by the total CMS energy and the particles’ masses:

2 22 222 2222 * sEE=()12 + =++ EE 1 2 22 EEpmm 12 = +++ 1 2 2()() pmpm + 1 + 2 ( p ≡p f )

22221 2 222 thu s: (p ++mpm12 )( )=−−4 ( s 2 p ( mm 12+ ))

4 22 2 221 2 4 222 2 22 222 p ++pm()12 m +=+mm12)44 () s p ++()42()4()mm12 − spsmm − 12 ++ pmm 12 + 11 solve for:ppsmmsmmmmsmmsmm222222222222222=++−+−=+−−+() ( ) 2( ) 4( ( ) 2( )) 44ss12 12 12 12 12 11λ(,sm22 , m ) =+= (smm−()()+ 2222*)( smm−− )≡→= ≡λ()( smm,,) →=p 12 44ss12 12 12 f CMS 2 s * 2 2 2 2 and similarly, |pi | = √λ(s,ma ,mb )/(2√s), with λ(s,ma ,mb ) the so‐called “triangle” function. Note that the triangle function is a Lorentz scalar!

10/19/2009 2 Laboratory System a + b → 1 + 2 Often the cross section is needed expressed in Laboratory system variables,

e.g. for electron‐proton scattering where the proton – the target b – is at rest, i.e. pb = (mb,0). 2 2 * 2 In the CMS: λ(s,ma ,mb ) = 4s|pi | . Expressed in Laboratory quantities (b at rest) it becomes: 22 2 2 2 2 22 2 2 λ()()()(sm,,ab m)()()=−( s m a + m b)( s − m a − m b) =+ s (()2() m a − m b )2() − s m a + m b

22 2 22222 22 =++(mmab 2 pp abab )( +−− mm )2( mm ab ++ 2 ppmm abab )( + )( p ab, 4-vect's)

22222222222 22 =+4(ppab ) 4( pp ab )( m a + m b )+()()2(2mmab++−−++ mm ab mm ab pp ab)(mmab+ ) 244222⎡⎤μ 222 =++−+=−4(ppab ) 2( mm ab ) 2( mm ab ) 4⎣⎦ ( pp ab; ) mm ab μ ⇓ In lblaboratory: =−()()2222Em mm Em += mm 4()4 m22 E −= m 2 m 22p ab ab ab ab b a a baLABSYS In order to find the elastic scattering cross section in the lab, we first express the differential cross section in Lorentz invariant quantities, and the scattering angle in terms of q2=t:

10/19/2009 3 Laboratory System a + b → 1 + 2 In order to find the elastic scattering cross section in the lab, we first express the differential cross section in Lorentz invariant quantities, and the scattering angle in terms of q2=t: mmb = 2 22 At the bqppppmppq ,2 vertex: −⋅ 2bbbbb =−− 2(22 ) ⋅ = 2 − 2 ⋅ = 2222 At the aqppmmEE,1 vertex: ≡− (()aaaa1111 ) =+++− 222+ 2pp cosθ

mmmmab==12,0 mm a =≈ 1 22 2 θ qmEEqEEEE=−2(ba −111 )LABSYS ≈− 2 a (1cos)4 −θ =− a sin LABSYS LABSYS 2 LABSYS mma= 1 mma= 1≈0 mmb= 2

22θ Ea solving for Eq1111 : =− 2( Eaba − Em ) =− 4 EE sin → E = LABSYS 2 2E 1sin+ a 2 mb 2 LABSYS mma= 1≈0 mmb= 2

mma= 1 θ mmb= 2 22* 222*** 2 θ q ≈−=−2(1)42(1ppcosθ )4si n CMS CMS 2 mma= 1 mmb= 2 i.e. for elastic scattering E1 depends only on θ ! 10/19/2009 4 IN‐elastic Scattering …

For inelastic scattering, where mb≠m2, it is a function of two independent variables,

• e.g. E1 and θ, 2 • or E1−Ea and the dimensionless variable “Bjorken‐x”: x ≡−2(pb⋅q)/q ; for elastic scattering x = 1, else 0 ≤ x < 1. It is now stihtftraightforwar d to express the LABSYS and CMS angles in terms of the appropriitate variables: 2 2 * 2 2 p * ** ⎛⎞EE11Ea dt==Ω==2cosp dθ d, dt22 mbba dE1 m E⎜⎟ d cosθ =−Ω d π ⎝⎠Eabm π from which we express the differential cross sections:

* dbdσ ()(a+→+b 12) 1 p f 2 *2*= mab+→+12 , dsΩ CMS 64π pi

dbdσ (12)1a+→+b ⎛⎞E1 2 = 22⎜⎟mab+→+12 dmEΩ LABSYS 64π ba⎝⎠ 10/19/2009 5 Example: the Decay a → 1 + 2 Again, a simple example is illustrative. Consider the decay of a partiilcle a into two fina l state partilicles 1 and 2 in the CMS of

particle a, where p1 + p2 = 0 and √s = Ea = ma: 4 (2π ) 1 ddpp123 2 Γ→+=(12)aEEm" δ ()pp12 +δ() 12 + −aam →+ 12 = 6 ∫∫6× 2(2)22ma π EE12

2 −1 11dp 2 pdpdΩ ⎡⎤∂+−()EEm 2 =+−=f δ()EEm12 a δ() pp − 22∫∫∫12 a mm ∫∫∫ ⎢⎥f 84ππmEEaa12 8 m 4 EEp 12 ⎣ ∂ ⎦ −1 2 11pdpdΩ ⎡⎤ p p 2 p 2 =+−=δ()pp f dΩ 2 ∫∫∫ ⎢⎥ f m 2 ∫∫ m 32ππmEEEEa 12⎣⎦ 1 2 32 mEa 12+ E

* p λ(,,)mmm222 Γ→+=(12)adf Ω**mm22 =a 12 d Ω 32ππ22mm∫∫aa→+12 64 23 ∫∫ →+ 12 aa()CMS

* 2 For a unity matrix element the decay width is simply Γ(a→1+2) = |pf |/(8π ma ), with dimension GeV−1; i.e. the matrix element must have dimension GeV.

10/19/2009 6 The Field Theories Local Gauge (Phase) Invariance is a powerful guiding principle for building theories and finding the form of the interactions Local Gauge Invariance is necessary for renormalizability of the Theory. All theories realized in nature are believed to be locally gauge invariant theories.

10/19/2009 7 Gauge Theories All true present‐day theories are based on local gauge (or phase) invariance, i.e. the phase of the wavefunctions can be varied according to an arbitrary space‐time dependent function.

The Lagrangian L=L(ψ,∂μψ) that describes the system keeps the same form under phase transformations of the type: ψψ()(rr,ttet)(→='(), )( ψitα (,)r () r, ) i.e. if ψ(r,t) is a solution, then ψ’(r,t) is a solution too! This phase transformation is called “local”; the phase depends on the local coordinates (r,t). Transformations with constant phase (i. e. the same for all space‐time points) are “global”. The term “gauge” is inherited from attempts by Hermann Weyl to construct theories with invariance under local scale (“eich”) transformations as a basis for electromagnetism. Local Gauge Invariance is a powerful guiding principle, and is necessary for renormalizability of the calculations. All theories realized in nature are locally gauge invariant theories. One can construct more complicated versions of the phase rotation, e.g. exp{iσ⋅b(x)} which is a two‐by‐two matrix, as can be seen from the series expansion of the exp function: 23 iσ⋅b ()()iiσ ⋅⋅b σ b ⎛⎞bbib312− ei=+1 σ ⋅+b + +... , with σ ⋅=b σσ11 bbb + 2 2 + σ 33 =⎜⎟ 2! 3! ⎝⎠bib12+− b 3 with the standard representation of the Pauli matrices. 10/19/2009 8 Massless Vector Bosons In the absence of sources and currents (jμ=0) the Coulomb potential V is zero (or constant), and equation ∂∂2A JK jA==∂∂−∂∂=jiiiμ AA02 −∇+∇ V μ 0 ∂tt2 ∂ Becomes the wave equation for free photons: ∇2A=∂2A/∂t2. With solutions: A(r,t) = a ε cos(k⋅r-ωt), with amplitude a and polarization vector ε. For the free photon the Coulomb gauge condition ∇⋅A=0 translates into k⋅ε=0 (or p⋅ε=0 with X), i.e. a free photon has only two independent components transverse to its direction! The fourvector field Aμ represents the photon, the carrier of electromagnetic interactions. Aμ is a four‐vector field: the photon is a spin‐1 (vector) boson. All force carriers are vector bosons (except for the spin‐2 Graviton). In preparation for the discussion of massive vector bosons W± (80 GeV) and Z0 (91 GeV), we consider the addition of a mass term to the Lagrangian: 11 μ μν 2 LFFmEM = −− μ42μν AA A gauge transformation of Aμ → Aμ’(x) = Aμ(x) + ∂μχ(x) is clearly not leaving this Lagrangian μ μ μ μ μ μ invariant:new terms like Aμ∂ χ and ∂ χ∂μχ are introduced by A → A ’(x) = A (x) + ∂ χ(x). Thus, gauge invariance requires the free photon to be massless. This is fine for electromagnetism, but is a problem for the weak vector bosons: if they also have to stay massless, then the theory is definitely not going to describe reality! 10/19/2009 9 The Klein‐Gordon Equation for a Complex Scalar Field A complex scalar field can be likened to a simultaneous description of two real scalar fields describing particles of precisely the same mass m. We will consider the symmetry that exists in the Lagrangian for such two equal‐mass fields: 11 Lm=∂∂()()μμφ μμφφ −22 φφφ + ∂∂ − m 22 2211 1 2 2 2

We could have started with a different set of fields φ1’ and φ2’, related to φ1 and φ2 by a rotation:

⎛⎞φφφφ1111''⎛⎞cos sin ⎛⎞ ⎛⎞ ⎛⎞ cos− sin ⎛⎞ ⎜⎟μμμ==⎜⎟ ⎜⎟and ⎜⎟ ⎜⎟ μμμ ⎜⎟ ⎝⎠φααφ2222' ⎝⎠−sinαα cos α ⎝⎠ α ⎝⎠ ⎝⎠ sin cos φααφ ⎝⎠' ⇓ 11 μμLm=∂∂+∂∂()()∂∂φφμμ +∂∂ φφ φ φ− 22 + 2 22μμ11 2 2 1 2 st μ μ 1term:('cos∂−∂−+∂+∂φαφαφαφα12 'sin) ('cos 12 'sin) ('sin 12 φαφα 'cos) ('sinφαφ12+ 'cos) α= 222 2 =cos''sin''sin''cos''α ∂∂+μμμφφ11 αφφ ∂∂+ 11 αφφ ∂∂+ 11 αφφ ∂∂=∂∂+∂∂ μμμφφ 11 φφ 11 '' 2 '' 2 nd 22 2 2 22 2 term:φφ12+= φ ( 1 'cos αφ − 2 'sin α φ ) + ( αφ 1 'sin + α 2 'cos φφ ) =+ 12 ' '

10/19/2009 10 The Klein‐Gordon Equation for a Complex Scalar Field We can also write the Lagrangian in terms of a complex scalar field φ with the identification:

φ =(= (φ1+iφ2)/√2.

The rotation (φ1,φ2) into (φ1’,φ2’) is equivalent to a global phase change: 11 1 φ '('')cossinsincos=+=φφiiieie φ αφ() + αφ − αφ + α φφ = φ−−iiαα ( += ) 2212 1 2 1 2 2 12 where we used the Euler relation eiα= cosα + isinα. The Lagrangian in terms of φ becomes: φ =+111(φφii ) φφ with φ , φφ real ⇒=− φ*** φφφ ( φφ) and: =+=− ( ),−i ( ) 222212 12 12 1 2 1 ⇒=∂∂+∂∂−Lm()()μμφφμμ φφ φ φ 22 + 2 2 ()11 2 2 1 2 μμμμ μ 22μμμ 1term:()()st 1 ∂+∂++φφ** φφ φφ φφ−i ∂ ()()− **∂ −= ( 22) μ ( ) μ =∂+∂+−∂−∂−1 ()()()()φφ** φφ φφ φφ φ φ ** =∂∂ * 2 () 22 11−i 2tnd erm:φ 2+=φφφ 2 () + *2 + () − *2 = φφφφφφφφ( ()() + *2 −− *2) = * 12( 22) ( ) 2 1 ⇒=∂∂−Lm()μ μ *2* 10/19/2009 2 11

φφ φφ The Klein‐Gordon Equation for a Complex Scalar Field 1 μ *2* Thus, the Lagrangian Lm=∂∂μφ φφφ − is invariant under φ → φ’ = e−iαφ. 2 ( ) Invariance of L under such a glbllobal symmetry transfiformation requires the existence of a conserved Noether current. We will derive this from first principles. For ease of derivation we’ll consider small phase rotations e−iα, with |α|<<1. Then: φφ''(1)=≈−=e−iα (1 φφδφiiα )φφδφ+ ,w δφ ith ≡−αφ δφφ and:φ '***==≈+=+ (ee−+iiααφφαφφδφ ) (1 i ) ***** , with ≡ i δφαφ Invariance of L requires: ∂∂LLμμ ∂∂ LL δL δφ==0 δφφφ + ( ∂μμ )( +δφ → δφφφ** ) = + ∂ ( )( + → ) = ∂∂∂μμ() ∂∂∂ () φφ φφ ∂∂LLμμ⎛⎞⎛⎞ ∂ L * (φ → φ*) indicates φδ δφ=+∂ φ δφ δφ ⎜⎟⎜⎟μ − ∂+→= ∂μ +→=() ∂∂∂∂∂φ ()φφ () terms with φ* ⎝⎠⎝⎠μμ replacing φ ⎡⎤∂∂LL⎛⎞ ⎛ ∂ L ⎞ * =−∂⎢⎥ +∂ +→=() ⎜⎟ δφ φ δφ ⎜ ⎟ ⎣⎦∂∂∂∂∂φφ⎝⎠() φ ⎝ ()α ⎠ μ μ ⎛⎞∂L * ⎡⎤∂∂LL* =−∂0(iαφφφ⎜⎟ +→)0=∂i μ * φφ− = ∂∂() ⎢∂∂∂()∂ φφ ()⎥ 10/19/2009 ⎝⎠μ ⎣⎦12 φ μ The Klein‐Gordon Equation for a Complex Scalar Field 1 The Lagrangian Lm=∂∂μφ μφφφ*2* − is invariant under φ → φ’ = e−iαφ. 2 () Invariance of L under such a glbllobal symmetry transfiformation requires the existence of a conserved Noether current:

μ ⎡⎤∂∂LL* δαLi= ∂ φ⎢⎥ φμμ* −=0 ⎣⎦∂()∂∂∂φ ()φ The final quantity in square brackets is a covariant fourvector, and is conserved:

μ ⎡ ∂LL* ∂ ⎤ ∂ jjiμμ=≡0ith0, with q ⎢⎥* φφ − ⎣∂∂()φφ ∂∂ ()⎦ where we added some constant “charge” q in the definition of the conserved current.

μμ

10/19/2009 13 Current Conservation The conserved four‐current for the complex scalar field Lagrangian becomes: * * jμ = iq[φ (∂μφ) − (∂μφ )φ]. This current has interesting properties: if the field φ represents a particle with mass m, then the field φ* represents another particle with exactly the same mass: it is natural to consider φ* to represent the antiparticle. * Indeed, under the interchange (φ ↔ φ ), the four‐current changes sign: jμ ↔−jμ indicating that the charge of the antiparticle must be opposite that of the particle. This true for a generalized charge: i.e. for every additive quantum number: strangeness, charm, etc. as well as for eltilectric charge. An easy way to find the conserved current is to start with the Klein‐Gordon equation(s): μμ22* ()0,()0∂∂μμ +mmφφ = ∂∂ + = and left‐multiply the first by φ* and the second by φ and subtract: φφ*2()0∂+m 2 = 22* φφ()0∂+m = *2 2* μ * μμ* φφφφ∂−∂ =∂⎣⎦⎡⎤((∂−∂)) =0, * * i.e. conservation of the four‐current jμ = iq[φ (∂μφ) − (∂μφ )φ]. 10/19/2009 14 φφ φφ Current Conservation in the Schrödinger Equation This is very much similar to what we did in QM with the Schrödinger equation to obtain the conservation of probability or probability current:

*2⎛⎞∂ 1 φφ⎜⎟++∇i =0 ⎝⎠∂tm2

⎛⎞∂ 1 2* φφ⎜⎟−+∇i =0 ⎝⎠∂tm2

⎛⎞***22*∂∂ 1 i⎜⎟φφφφ−+ φφφφ( ∇∇∇ − ∇ )= 0 ⎝⎠∂∂tt2 m ∂ 1 JK JK JK ()φφ φ*** φ φ+∇⋅∇−∇= φ ( ) 0 ∂tmi2 ∂−JKi JK JK ⇒+∇⋅=≡ρ j 0 with ρφφ*** , φj ≡ φ φφ() ( ∇−∇ ) ( ) ∂tm2

10/19/2009 15 Conserved Currents * * K‐G current: jμ = iq[φ (∂μφ) − (∂μφ )φ] 1 JKJK Schdihrödinger current: j ≡−i φφ**()() ∇∇∇ − φφ∇ , ρ ≡ φ*φ 2m ( ) However, there is a fundamental difference between the “Schrödinger” current and the Klein‐ Gordon current: ρ = j0 has to be positive definite in order to be identified with a probability density; negative probabilities do not make sense. In the Schrödinger current this condition is satisfied by the identification ρ ≡ φ*φ = |φ|2=j0. 0*00* In the K‐G equation we find: ji=∂−∂⎣⎦⎡⎤φφ()() φφ ⇓ using solutions: φ ==ae−⋅−ipx ae i()pr Et jiaiEaE0 =−=22( 2 ) 2 (hence: a = 1 2 E ) where the parameter E is not positive‐definite because E = ±√√(p2+m2). Formally, solutions of the type φ = ae−ipx with negative E cannot simply be ignored because the set of base vectors spanning the Hilbert space would be incomplete. Because of the negative energy solutions, the Klein‐Gordon equation (which was considered by Schrödinger and later by Klein and Gordon) was first abandoned in favor of the Schrödinger equation. 10/19/2009 16 Particle and Anti‐Particles Later still, after the Dirac equation became firmly established and when Dirac successfully tamed its negative energy solutions, the Klein‐Gordon equation started a second life. Ultimately, Ernst Stueckelberg and later Richard Feynman re‐interpreted the negative‐energy particles going forward in time as positive‐energy antiparticles going backwards in time (i.e. reversing direction): ee−−−−ipx=⇔−−⋅iEt( pr) e −−−−−⋅− i(()()()() E t p r ) = e i()() p x * * We can consider what happens to the charge current : jμ = iq[φ (∂μφ) − (∂μφ )φ]: j0 represents a charge density (instead of a ppyrobability density) and may be positive or negative.

For electrons of positive energy E1 or negative energy –E1, and charge –e we have: ⎧ 0 2 μ 2 μ ⎪ jeaE==<−201 < electron with EE=>11 0,pp = : j =− 2 eap , and: ⎨ 2 μμ⎩⎪ jp=−2ea 1 2 electron with EE=−11 < 0,pp = : j =− 2 eap , 22 ⎪⎪⎧⎫jeaEeaE0 =−2()2 − =+ > 0positron with positive energy and: ⎨⎬11= 22 and traveling backwards! ⎪⎪jp=−22()ea11 =+ ea − p 10/19/2009 ⎩⎭ 17 Local Gauge Invariance Above, we saw the invariance of the complex scalar field Lagrangian under phase transformations of the type φ → φ’ = eiαφ, with α a constant. This is a so‐called “global” gauge (phase) transformation. This means that we require that the fields all phase‐rotate together everywhere and for all time. This doesn’ t seem very elegant: imagine that there is no electromagnetic interaction and only the strong interaction existed;

in that situation the proton φ1 and neutron φ2 would be indistinguishable and have the same mass, and any mixture of the two would be simpyply the same nucleon. Someone in the US and someone in China might define different mixtures as their “nucleon” but we would expect that the same (strong interaction) theory would work well in either case. Let us investigate what would happen to the Lagrangian and the equations of motion if we were to require invariance under local phase transformations: φ()x →=φφ '()xeixα () () x, where x, as usual, is the space‐time coordinate xμ.

10/19/2009 18 Local Gauge Invariance Local gauge invariance : i.e. invariance of the Lagrangian under local phase transformations: φφ()x →=' ()xe φixα () () x, where x is the space‐time coordinate xμ. Of course we expect trouble in the Lagrangian: after a local phase transformation of the fields the derivative terms will act on both α(x) and φ(x), thereby leading to extra terms involving the scalar function α(x). Otherwise stated: the local phase factor does not simply “pass through” the differentiation operator: μαix μ() α α μ α μ μ α μ ix () ix () ix () ix () ∂=∂+∂=∂+∂≠∂(exiφαφ()) ( () xexe) ()( () φxeix) ( αφ( ()) ) () xe φ( () x)

i.e. ∂μφ transforms differently from φ itself, which is the cause of the non‐invariance: if they both would be transforming in the same way, we could simppyly divide out the local phase factor in the‐Euler Lagrange equation.

10/19/2009 19 Local Gauge Invariance As an example, consider the difference between a global and a local phase transformation on * * the conserved current, jμ = iq[φ (∂μφ) − (∂μφ )φ]: μμμ⎡⎤**⎡ −−ii αμαμαα * ii *⎤ Global phase α jiq '=∂−∂=⎣⎦φφ ' ( ') ( φφ ' ) ' φ iqe⎣ φ φ( ∂ ( φ e )) −∂( ( e )) e⎦ = −−ii**** i i = iq⎣⎦⎣⎦⎡⎤⎡⎤ eφφ e()∂∂∂− e φφφφφφ (∂ ) e= iq ()() ∂∂∂ − ∂ = j ααμ αμ α μ μ μ μμμ⎡⎤**⎡⎤−−ii αμαμαα * ii * Local phase αφφφφφφφφ (x ) j '=∂−∂= iq⎣⎦ ' ( ') ( ' ) ' iq⎣⎦ e( ∂ ( e )) −∂( ( e )) e = ⎡⎤−−−iii*** iii =∂+∂=∂+∂∂+∂=iq⎣⎦ eφαφ( i() e e () φ)−−( αφ i ()∂+∂= e φφ e ( )) e αμααμ μα αμα −−iiααμ*** αμ i α i μ μ μ * μ =∂−∂−∂=iq⎡⎤⎣⎦ejφφ e() e φφ ( ) e 2( q αφφ) −2q()∂ αφφ≠ j In order to implement local gauge invariance of the Lagrangian, we will have to add cancellation terms in the Lagrangian that, under transformations, acquire additive terms that can cancel the underlined term. This reminds us of the gauge freedom of the photon field Aμ: the physics was unchanged if we shifted Aμ by the derivative of a scalar function of space‐time, very much like the ∂μα term that we need to cancel here …

10/19/2009 20 Gauge Freedom of Aμ and Local Gauge Invariance Let us try to see how the combination (∂μ+Aμ)φ transforms: μμixαμμ() μμ φφ()x →= '()xeφμμ () x,an d: AAA ()x →A '()x =A ()x −∂ χ μμ iiiαα ()∂+Aiφαχ→+(∂∂+A )''φ ==∂+ee()i()∂ φ + ()A − ()∂ eA( + ()() ∂ −∂ φ ) μ i.e. almost success! χ The replacement that works is theα identification χ=iαμ, or the combination (adding q!): μ μμ μμμ μμμ μμμ ∂→∂+iqAμμ (with A αμ' = A −∂α / q μ ), or: μμ i ∂→∂− i α qAα or: p → p − qA which is the ppprinciple of minimal replacement (W. Pauli) gggiving the EMμμ interaction: μ ixαμμμμ() φ Simultaneous transformations:φαφ φ (x )→=φφ '( x ) e ( x ),α Ax ( ) → αφ A '( x ) = Ax ( ) −∂μ q ii ()''∂+iqA = e() i() ∂ +∂ + iq() A −∂() q e =

iiαμ μ μ μ αμ μ =∂++∂−∂=∂+eiqAii( ( ααφ)() ) eiqA( φ) Thus, the combination (∂μ+iqAμ)φ transforms like φ itself! and the replacement ∂μφ →(∂μ+iqAμ)φ everywhere in the Lagrangian will make it invariant under local phase transformations of the field φ → φ’ = e iαφ, and when we use the gauge freedom of the electromagnetic field to choose to simultaneously shift Aμ → Aμ’ = Aμ −∂μα/q. 10/19/2009 21 The Covariant Derivative The combination (∂μ+iqAμ) is defined as the covariant derivative denoted as Dμ ≡ ∂μ + iqAμ. As example we show that the Klein‐Gordon current in is indeed invariant with the replacement of ∂μ by Dμ: μ ⎡⎤***μμ ααμαμα⎡⎤−ii− i * i j ''('')('')'=−=iq⎣⎦φ Dφφφ D iq⎣ e( e φφ ()() D) −( e D φφ) e ⎦ = φφ φφ ⎡⎤**μμ μ =−=iq⎣⎦()() D D j In summary: requiring local gauge invariance of the Lagrangian forces the introduction of a “counter term”, involving a vector field that exhibits the correct form of the (electromagnetic) interaction! The elliectromagnetic fie ld and its iiinteractions is generated by the gauge priiilnciple, the freedom of choosing a phase locally for the particle fields.

This is very reminiscent of the theory of general relativity, where the form of the gravitational interaction follows directly from the freedom to choose my coordinate system locally, and no universal inertial system is required. 10/19/2009 22 General Local Gauge Invariance Often the local phase function exp{iα(x)} is written as exp{iqα(x)}, which amounts to a redefinition of the space‐time dependent function α(x) → qα(x). The parameter q can be thought of as the eigenvalue of the charge operator Q, and, because the phase factor may be thought of as an operator itself, it can also be written as exp{iQα(x)}. This will become useful later. We may generalize the local gauge transformations, and thus allow the gauge principle to prescribe the other types of interactions in nature: the weak and the strong interactions.

A general local phase transformation can be written as: φ(x )→=φφ (xUxx )' ( ) ( ), with covariant derivative: Dμ =∂+μμ igB

U(x) must be a unitary transformation, because it should respect the normalization of the fields.

10/19/2009 23 Transformation Properties of the Gauge Fields A general local phase transformation can be written as: φφ(()xxUxx )→∂+→ ()'( )'== φ (()() ) ( ), with covariant derivative: Dμμ∂+ igB μ with U(x) a unitary transformation. We can then derive the transformation property for Bμ that allows the Lagrangian to be invariant when the covariant derivative replaces the “normal” derivative ∂μ everywhere: invariance requires:μφφ μ (DUDμμ φφ )' φ= ( φ μμ ) φ μμ require ⇒==∂+=∂+=∂+()'''(D D igB ')( U igB ') U U ( igB ) ⇒ (')()∂∂∂μμ+=iBigBU Uφφ μμ U∂ + iiBgB solve for Bμμ ': igB ' Uφ =−∂ μ ( U μφφ ) + U ( μ ∂ ) + igUB μ φ μ =−∂ ( φ μ U ) + μ igUB φ φ =() −∂ ( U ) + igUB i i ⇒=BU'μ ( ∂+μμUUB))() ⇒ μμ B' = UBU−−11 μ +∂() UU g g

Note that we need to keep the order of operations because U may be a transformation matrix, for instance U(x)=exp{i½σ⋅b(x)}, and Bμ may be a vector or tensor. To check, try U=eia: AUAUiμμ'=+∂−−11 μ1111αμαμα () UU with UeAeAei ==iiiiii :' − +∂=+∂=−∂ μααμ ( eeAiiee ) − ()α − αμαμμ i A ()α qqqq 10/19/2009 24 Consequences of Local Gauge Invariance Consider the Klein‐Gordon equation for a bosonic field φ, (e.g. the π+); the complex conjugate field φ* then represents the anti‐π+, iei.e. the π−. + We require L toμμ be invariant under local phase transformations of the π field (q=+e), which forces theμμ substitution ∂μ → Dμ ≡∂μ + ieAμ. Thus: ⎡⎤22 ⎡μμ 2 ⎤ KKG-Gequation equation ⎣⎦∂+mieAieAmφφ → ⎣()() ( ∂+μ μ )( ∂+ ) + ⎦= 2 222 22 22 =∂+⎣⎦⎣⎦⎡⎤⎡⎤ieAA( ∂ + ∂ ) − eAm +φφφ →∂+ m =− V , with: VieAA ≡ ( ∂ + ∂ ) − eA Where potential V contains the interactions: a term of 1st order in e, and a 2nd order term. We will assume that V vanishes at large distances from the “interaction region”. st (1) The 1 order transition amplitude Tfi from initial state φi to final state φf is thenμμ the integral: μμ TidxxVxedxxAAxedxA(1)=− 4φφ *() () =+ φ 4 φ * ( φφφφ )( ∂μμ + ∂ ) μ () =+ μ 4⎡ * ∂ ( ) + * A ( ∂ )⎤ fifi∫∫ fμμ μ i μ ∫⎣⎦ fifi =+edx4**⎡⎤⎡⎤⎡⎤ −∂(μμφφφ ) Aμμ + A () ∂ μ φ μ + φφ e * A μμ φφ =− idxie φφ 4* ()() ∂ − ∂ * A + 0 ∫∫⎣⎦⎣⎦⎣⎦fif i fiS fi fi μ =−idx4 jjjfiA , with: j fi ≡ie ⎡⎤φφ**()()∂ −∂ φφ ∫ μμ⎣⎦fiμμ fi −ipx –½ The φi and φf are the free particle wavefunctions ae , a=(2E) , because they are defined (prepared or measured) far away from the interaction region both in time and in space. 10/19/2009 25 The Current‐Current Interaction The interacting current can then be written as (with φ =ae−ipx): ip x− ip x 1 ifμ fi⎡⎤⎡⎤⎡⎤*** i ffi f i −ip()μ − pμ x jμμμ= ie⎣⎦⎣⎦⎣⎦φφfi()()∂∂∂ − φφ∂ fifi=−− iea a μμ ip ip e μμ = e p + p e 22EEfi μ The interaction can be thought of as a pion current jμ , which emits or absorbs a photon A . The photon can be thought of as being “sinked” or “sourced” by second current, ege.g. by a positively charged Kaon K+. In order to prevent (?) confusion we will number the participating particles –the pion and the Kaon – as follows: the incoming (initial state) particles are 1 (pion ) and 2 (),(Kaon), and the outgoing particles are 3 (pion) and 4 (Kaon). The exchanged photon remains unnumbered. μμμμμ K + πThe exchanged photon has a four‐momentum equal to q = p3−p1. π + q ≡ p3−p1

The pion and Kaon currents are then written (see equation ) as: p4 p3 q 11μμ −−ip()13 p x −−ip()24 p x γ jje==13[] ppe 1 + 3; jjeK == 24[] ppe 2 + 4 , p2 22EE 22 EE p1 13 24 interaction volume, t + i.e. region in space- π + K with pp13− =− pp 2 + 4 = q x time where Aμ≠0. μ μ μ Note that Current Conservation requires: ∂μ j = 0, thus: qμ (p1+p3) = qμ (p2+p4) = 0; this easily verified: q (p +p )μ =(p –p ) (p +p )μ = m 2–m 2 =0! 10/19/his 2009 μ 1 3 1 3 μ 1 3 1 3 26 The Current‐Current Interaction μν ν The photon isνμμ related to its source/sink by the inhomogeneous Maxwell equations: ∂μF =j . μ μ We use the Lorentz gauge‐fixing condition ∂μA =0 to obtain a simple equation for A : νμννμννμ ν ν μ ν μ ν μν 2 2 ∂μμFAAA =∂( ∂ −∂ ) =∂ −∂ ( ∂A ) =∂=Aj (in the Lorentz gauge) nd μμμ Solving this 2 orderμ differential equation for A : μμ 2 11−−ip()24 p x iqx ∂===Aμμ j j e[ ppe +] = e[ ppe +] μ K 24 24 24 22EE24 22 EE 24 1 μ −1 −μ2 iqx ⇒=Aiqe() [] ppe24 + =2 j 24 22EE24 q νν st (1) ν The 1 order transition amplitude Tfi becomes: −1 −gμν T(1)=− idxj 4 13Ajj =− idxj 4 13 =−= idx 4 jμ fi ∫∫q2 24 ∫ 13 q2 24

1 −gμν =−idxeep()+ pν ep (+ p)μ 4 −+−ip()12 p p3 − px 4 = 132 24∫ 22EE1234 22 EE q ν 1 ⎛−⎛⎞igμν ⎞ =−i dx4 e−+−ip()12 p p3 − px4 ×() −i ie() p++ p ie ( p p )= ∫ ⎜ 13⎜⎟2 24⎟ 22EE1234 22 EE ⎝⎠q ⎝ μ ⎠ 4 1 ⎛⎞−ig =−i (2πδ )4 ()pppp+ −− ×(),−i with −=i iep() + pν μν iep()+ p μ ∏ 123 4 mfi mfi 13⎜⎟2 24 j=1 2E q 10/19/2009 j ⎝ ⎠ 27 The Current‐Currentνμ Interaction 1 ⎛−⎛⎞igμν ⎞ T(1) =− idxe4 −+−−ip()1234 p p px ×() − i iep ( + p ) iep ( + p ) fi ∫ ⎜⎟13⎜⎟2 24 22EE123 22 EE4 ⎝⎠⎝⎠q ν 4 1 ⎛⎞−ig = −×i (2πδ )4 (p +−−p pp) (−i ), with − i=+ iep(( p) μν iep + p)μ ∏ 1234 mfi mfi 1243 ⎜ 2 ⎟ j=1 2E j ⎝ q ⎠ st where the 1 order single‐boson‐exchange matrix element mfi has the universal form:

−imfi = (vertex factor)13×(photon propagator)× (vertex factor)24. The vertex factor differs for bosons and fermions, and the propagator differs for a gluon or for a weak vector boson, but the structure of the matrix element, and the “normalization” factors in front remain unchanged. The delta distribution function represents energy‐momentum conservation, and arises naturally from the integration of the exponentials in the free particle plane waves. We have now derived the simplest Feynman rules for QED: 2 • a photon exchange gives rise to the factor −igμν/q (in the Lorentz gauge!) μ • a scalar1‐boson‐scalar3 vertex contributes a factor ie(p1+p3) , where e is the charge and the momenta refer to the incoming and outgoing particles. Using the Feynman rules, we can simply and efficiently calculate all graphs that contribute to a particular process. 10/19/2009 28 ⎯ Example of Toplogical Relations For example, the muon decay diagram (a) is related to the exchange diagram (c) by a μ μ simple “flip” p2 → −p2 of particle 2’s momentum in the amplitude (!) and by the replacement of the particle by its backwards going antiparticle (note the arrows!):

e− e+ q ≡ p −p νe ⎯νe 3 1 ⎯νe νμ ν ν p2 μ p2 μ p p p p q 4 p q 4 p q 4 3 ==3 3 W W W −p2 p1 p1 p1 t t t e+ − − − μ x μ x μ x (a)(b)(c)

10/19/2009 29 Example: Scattering of Spin‐0 Particles: a+b→a+b We calculate the electromagnetic interaction of “spinless” e+ and e– at 1st order in e:

These are expressed in terms of the matrix element mfi. This matrix element squared will be a Lorentz scalar, and we’ll express it in terms of Lorentz invariant quantities, e.g. in terms of the Mandelstam variables 2 2 2 2 s≡(p1+p2) =m1 +m2 +2p1p2=(p3+p4) , 2 2 2 2 2 2 2 2 t≡(p1–p3) =m1 +m3 –2p1p3=(p2–p4) , and u≡(p1–p4) =m1 +m4 –2p1p4=(p2–p3) .

Similar expressions can be formed using energy‐momentum conservation: p1+p2= p3+p4. Spinless e– μ– → e– μ–; m =m , m =m : νμ1 3 2 4 ⎛⎞−−++ig ep2 ()() p p pμ −=−+iiepp⎡⎤⎡⎤() μ μν − ieppi ( + ) =− 13 24 mfi ⎣⎦⎣⎦13⎜⎟22 24 qpp()13− – ⎝⎠ – μ This can be expressed silimply in terms of s, t and u: e 22μ q≡p1−p3 p4 eeμ p3 −=im i()() pp + pp + = i() pppppppp + + + q fi qq2213 24 12143234 γ p2 mmmm==;;⇒ pp= pp pp= pp, 1324 12341423 p1 – e2 e2 e – t μ thus: − iiusm =+=ipppp() 2 2 − ()− x fi q2 12 14 t 10/19/2009 30 Example: Spinless e– e+ → e– e+ There are now two contributing diagrams: 1. a “t‐channel” exchange diagram (as before) 2. an annihilation diagram or “s‐channel” diagram + – Both amplitudes must be added together! e – e e – e p q≡p1−p3 p4 p q≡p1−p3 -p4 We use m1=m2=m3=m4 in the last line: 3 q 3 q

2 μ γ p2 = γ -p2 −+−−ep()()13 pμ p 24 p p1 p1 −=−iimfi 2 t + t – ()pp− e – e e – e 13 x x + +

2 μ – + – – −−−+ep()()12 pμ p 43 p e e e e − i p4 -p4 2 p3 p3 ()pp12+ t q ≡p1+p2 t q ≡p1+p2 γ = γ s − utu− x x 2 ⎡⎤p1 p2 p1 -p2 =−ie() − − − e – e + e – e – ⎣⎦⎢⎥ts 10/19/2009 31 Example: Spinless e– e– → e– e– Again, there are two contributing diagrams: Both are “t‐chl”hannel” exchange diagrams; they differ because the fina l state partiilcles (spinless bosons!) are indistinguishable. – Again, both amplitudes must be added together: e – e p q≡p1−p3 p4 3 q γ p −+ep2 ()() p p + pμ 2 13μ 24 p1 −=−iim t – fi 2 e – e ()p13− p x + + – q≡p −p – 2 μ e 1 4 e −+ep()()14 pμ p 23 + p p p4 − i 3 ()pp− 2 14 γ p q 2 2 ⎡⎤su− st− p1 =−ie() − + t – ⎢⎥ e – e ⎣⎦tu x 10/19/2009 32 The Real Pion: Form Factors… The real pion is built from quarks, and therefore will not have a point‐like EM interaction. μ Although the current jπ of the real pion will differ from the point‐like “pp,ion”, it must still be Lorentzμμ contravariant vector!

At the scalar‐photon‐scalar vertex only 3 fourvectors are available: p1, p3, and q=p1−p3, of μ μ μ μ μ which only 2 are independent; choose: q =(p1−p3) and (p1+p3) . Note that q ⊥ (p1+p3) . 2 2 2 2 Fina lly, there are only two Lorentz scalars at the vertex: (p1) = (p3) = mπ , and q = (p1−p3) = μμ2 2 π μμμ μ 2μπmπ μ −2 p1p3 of which only one is independent, choose q . The most general current fourvector that can be constructed for the real pion is then: e μ ⎡⎤22−−ip()13 p x jjπ ==13⎣⎦ FqppGqqe()( 1 + 3 ) + () 22EE13 Current conservation further restricts this to a single unknown Form Factor F(q2): μμμ ee⎡⎤⎡⎤22iqμ x 22−iqx 0=∂j =∂j13 =⎣⎦⎣⎦ Fqp ( )( 1 + p 3 ) + Gqq ( ) ∂ e = Fqp ( )( 1 + p 3 ) + Gqqqe ( ) 22EE13 4 EE 13  =0  μμμ 2222μμμ 22 ⇒++=⇒=+=−+=+=Fqqpp()()()0()013 Gqq Gq b.c.: qpp()()() 13 pp 13 pp 13 p 1 p 3 0 μ e 2 μ −iqx Thus: j"l""real π "= Fq()( p13+ p ) e 22EE13 F(q2) is normalized to 1 at q2=0, because e is defined at large distance (zero momentum transfer q), i.e. at large distances the pion again appears point‐like. 10/19/2009 33 Massive Vector Bosons Free massive spin‐1 bosons are described by the free Klein‐Gordon equation: (∂22+=MW) μ 0 In the restframe of the Wμ the spin can be characterized by a polarization 3‐vector ε, ege.g. expressed in the basis of linear polarization vectors εx=(1,0,0), εy=(0,1,0), εz=(0,0,1), or as circular and longitudinal polarization states ε1=(εx + iεy)/√2, ε2=(εx –iεy)/√2, ε3= εz . Under Lorentz transformations, the polarization vector transforms as the 3‐vector part of a fourvector εμ=(0,ε) in the restframe. In the boson’s restframe we trivially have μ μ ν pμε =(M,0) gμν(0,ε) = 0, which, being a Lorentz scalar, will remain true in any frame. Hence, a massive spin‐1 object has only three independent polarization states, chosen, for instance, as 2 transverse and 1 longitudinal state. Note, that the notions transverse/longitudinal are not invariant! The K‐G equation for the massive vector field Wμ above was first derived by Alexandru Proca (1938) from the the massless photon Lagrangian by the substitution ∂2→∂2+M2: LFFLFFMWWFWW μ =− μν 111μν μνμνμμνμννμ → =− −2 ,with ≡∂ −∂ EM 442 The Euler‐Lagrange equations of motion for Wμ are: 22μμ ν μνμν 22 (∂ +MW) −∂ ∂ν W ={( ∂ + M) g −∂ ∂} Wν =0 i.e. four independent equations.

10/19/2009 34 Massive Vector Bosons As an important topic in QED, we discuss Vector Bosons, i.e. bosons with spin‐1, massive or not, that are carriers of the electromagnetic interaction. Lagrange equation for a massive m=M vector field Wμ: LFFLFFMWWFWW μ =− μν 111μν μνμνμμνμννμ → =− −2 , with ≡∂ −∂ EM 442 The (four) Euler‐Lagrange equations of motion for Wμ are: 22μμ ν μνμν 22 ()∂ +MW −∂ ∂ W ={( ∂ + M) g −∂ ∂} W =0 νν 2 μ μ 2 Acting on equation with ∂μ from the left yields M ∂μW =0, iei.e. ∂μW =0 if and only if M ≠0! So, if M2≠0 then (∂+22MW) μ =0 is true as four independent equations with one constraint. μ Note, that the constraint ∂μW =0 directly results from the Lagrangian, not as a gauge choice, but only so if M2≠0. μ In 4‐momentum space, the constraint corresponds to pμε = 0 when we write the solution for the free field as Wμ=εμe–ipx.

10/19/2009 35 The Propagator of the Massive Vector Boson The propagator equation emerges when we add a point‐source δ4 to the right‐hand side of the free‐boson equation : 22μν μ ν 4 {}()∂+Mg −∂∂ S =δ () xx − ' −−ip(') x x 4 e with formal solution: Sxxμν( , ')= dp μν ∫ 22μν μ ν (− p ++Mg) pp −1 δδ22μν μ ν The Fourier transform of S(x,x’) is Sp()=−+{( p M ) g + pp} . This seems an “inverse” operator; let’s find its “normal” form: 22 μ μν μ ν {()−+pMgμμμ + pp}{Ag+ Bp p } =δ , with functions A and B to be determined... ρρρ ⇒−+++(pMA22 )() AM 2Bp p =⇒−+=+= ( p 22 M )1 A and ()0 A MB 2 νρ νρ μμμν μ ν 2 1 −1 μν −+gppM+ ⇒=AB22; = 222 ⇒Sp()=− i 22 −+pM M() − pM+ ρ pM− Vector Boson Propagator

S(p) is then the massive vector boson propagator. When the propagator momentum is much smaller than the boson mass M: −ig μνμνig Sq()≈≈ 2 22qM22 ()qM− W W MW 10/19/2009 36 Massless Vector Bosons –the Photon μ μ For the free photon with momentum k no restframe exists and therefore the constraint kμε μ = 0 or ∂μA =0 is just a choice we make; other gauge choices are possible. μ This Lorentz gauge ∂μA =0 is often chosen when dealing with tree‐level diagrams, as here. The gauge choice reduces the 4 equations ∂2Aμ=0, to 3 independent ones. For the free phthoton, this does not exhhtaust all ffdreedom yet; i.e. we still have more ffdreedom to μ choose A withoutJK affecting the physical observables: JK JK BA=∇× and E =−∇00 A −∇A (EFii = 000 =∇ A −∇ A i ) It is siilmple to chhkeck that any scalar fie ld Λ()(x) wihith property ∂2Λ=0 preserves bhboth the Lorentz condition, as well as the physical fields B and E. This extra gauge freedom allows yet another constraint on the εμ components, which are therefore reduced to only 2 independent ones, typically chosen transverse: choose Λ=iikxα exp( − ) ⇒ ∂μμ Λ=ααkexp() − ikx (Note: ∂22 Λ= kexp ikx = 0 because () k 2 =0) then A 'μ =+∂Λ=+Akikxikxkμμ(εα μμ ε ε εα) exp μ μμμ() − = ' exp − () , with ' =+ ,

10/19/2009 37 Polarizations of the Free Photon choose Λ=iikxα exp( − ) ⇒ ∂μμ Λ=ααkexp() − ikx (Note: ∂22 Λ= kexp ikx = 0 because () k 2 =0)

μ μμ then AA '=+∂Λ=+μμμ() μ μ μ k exp() − ikx = ' exp− ikx () , with ' =+k , The last expression, the arbitrary “shift” freedom, is an expression of the gauge freedom in ε QED; αε it indicates that εμ and ε’μ, representing the same photon, are related by an arbitrary α times the ph’hoton’s four momentum, and thus we may always choose α such that ε0 = 0 e.g. suppose that ε0 ≠ 0, then choose α = –ε0/k0, then ε0 = ε0 + αk0 = 0. Therefore, with this choice of gauge, the Coulomb gauge, the free photon isεεα fully transverse: μ εμ k = ε·k =0.

These two remaining independent components can be chosen as εx and εy, or as left and right‐circular polarizations.

10/19/2009 38 Interacting Photons –Principle of Minimal EM Interaction From the free‐particle equations of motion (∂2+m2)ψ=0 we went to the situation where photons interact with particles by requiring local gauge invariance of the free Lagrangian. This led to the “principle of minimal EM interaction”, which states that by making the replacement ∂μ → ∂μ + ieAμ (with e the particle’s charge) we obtain the correct form of the coupling: 0 =∂+⎡⎤⎡⎤ieA ∂+μμ ieA μ + m μ2222ψψ =∂∂+∂ μ ie A +∂ A− e A+ m ⎣⎦⎣⎦( μ μμμμ)( ) ( ) ()∂+22mieAAeAψψ =−⎡⎤ ∂ + ∂ ( + 22 )  ⎣⎦ μμ scalarμμ field source! μ μ –ikx The amplitude for emission of a photon A =ε e by a particle with momentum p1 before emission, and momentum p after emission, is: 2 μμ μ μ Tdx()ee==−∂+∂=−−∂+∂ 4ψψ *V () ψ ψ iedxAA ψψψψ 4 *μ μμ iedx 4⎡ ()() * * ⎤ A 21∫∫ 2 1 2( ) 1 ∫⎣ 1 2 2 1 ⎦ VT VT VT μ 4 =−iiep⎡⎤() + pμε NN dx exp() − ip − p + kx ⎣⎦12 12∫ [] 12 VT =−ii⎡⎤ep() + p ε μ NN 2π 44δ (pp − +k ),  ⎣⎦12μ 12 12 mfi –½ 10/19/2009with the normalizations as before: Nj ≡ (2Ej) . 39 Interacting Photons The interaction term quadratic in e gives:

22μ TdxedxAA()ee== 4ψψ *V () ψ 2 ψ 4 * (3)(4)μ 21∫∫ 2 1 2μ 1 VT VT =−−++eNNNNdxippkkx2(3)(4)εε 4 exp ( ) μ 12333412∫ [ 3 4 ] VT 2(3)(4)μ μ 4 4 =−++eNNNNppkkεε πδ 1234()(2 1 2 3 4 ) and with proper symmetrization because 3 and 4 are indistinguishableindistinguishable: :

2 4 Tiie()e =−⎡⎤ 2()2εε (3)(4)μμ εε πδμμ + (4)(3) NNNNppkk()( 4 − + + ) 21 ⎣⎦ 1 2 3 4 1 2 3 4 mfi

10/19/2009 40 The Propagator of the Spin‐0 Boson In equation we derived the expression for the photon propagator from the inhomogeneous Maxwell equations (the Euler Lagrange equations of the EM Lagrangian). We follow a similar argumentation in deriving the propagator of a massive spin‐zero particle. The non‐free Klein‐Gordon equation is: (∂22+=m )φ J()x Again, we choose theφδ source J(x) as a point source in space‐time: −−11e−iqx (∂+ φ22mx φ )φδ === 4(() ), which has formal solution () ( x ) dqq4 ; ()( ) (2π )42222∫ qm−− qm 22−iqx −+11()()−+iq m e check: (∂+22m) () x = dq 4 = dqe 4−iqx = 4 () x (2π )4224∫∫qm− (2π ) where we identify the solution φ(q), the Fourier transform of φ(x), as the pion propagator. Without proof: for consistency with higher‐order loop‐diagrams, a factor –i must be added to this propagator.

10/19/2009 41 Example: Bremsstrahlung by a Scalar Boson Bremsstrahlung occurs when an electron is accelerated in a nuclear Coulomb field and emits a photon, while Compton scattering is the process of X‐ray photon scattering off quasi‐free electrons in matter. Both processes are pure QED and calculable (spinless!) with the tools we developed above. Take a spin‐0 charged pion scattering in the Coulomb field of a heavy nucleus of charge Ze. We count three diagrams; in the limit of very large nuclear mass M, we can ignore the diagrams in which the photon radiates off the “heavy” lines p1 and p3. Note that the third diagram is really two because of the interchangeability/symmetrization of the two photons, iie.e. there is an extra factor two in the matrix element for (3). We simplify the formulae using transversality of the emitted real photon and assume the nucleus to be very massive (this also suppresses brehmsstrahlung off p1 and p3!).

1 2 3 Z, p Z, p Z, p3 3 3 Z, p1 Z, p Z, p q 1 1 p4-q x q k, ε q k, ε =p= p2-k↓ t

p2+q↑ + + + = p4+k + + + π p2 k, ε π p4 π p2 π p4 π p2 π p4 10/19/2009 42 Bremsstrahlung 1 2 3 Z, p Z, p Z, p3 3 3 Z, p1 Z, p Z, p q 1 1 p4-q x q k, ε q k, ε = p2-k↓ t

p2+q↑ + + + = p4+k + + + π p2 k, ε π p4 π p2 π p4 π p2 π p4 The matrix elements describing these Feynman diagrams are: μν 3 (1) μν−−ig i ρμ2 ie 2 p2⋅ε −=+imfi iep()()13 p2222 iep 4 +− p 4 q iep () 22 +−= p kε () p 13 + p p 4 qpkmqpk()2− −−⋅ 22 μνμν μ 3 (2) μν−−ig i ρμ2 ie 2 p4⋅ε −=+iieppieppqmfi ()132222 ( 22 ++ ) iepkp ( 4 ++= 4 )ε () ppp 13 + 2 qpkmqpk()4+ −⋅ 24 −iiig μν +2ie3 −=+iiepm(3) ( piepp)2()2 εε=+μ ρμ fi 1 313qq22 where we used the following equalities: μ 22 2 2ρμ kk⋅=εεμ =0, ( pkmmpkmjjj±±±+± )−=± 2⋅ + 0−=± 2 pkj⋅= , 2, 4 μμ22 2 2 μμ (ppq13+=+−=−=−= ) ( pp 13 ) ( pp 13 ) p 1 p 3 M M 0 (Current Conservation!)

10/19/2009 ⇒+()(2)2(),2,4pp13 pqjj ±=+ ppp 13 j = 43

μμ μμ Example: Bremsstrahlung by a Scalar Boson The matrix elements describing −2ie3 2 p ⋅ε −=ippp(1) () + μ 2 the three Feynman diagrams are: mfi 2 13μ 4 qp−2 2⋅k 3 (2) −2ie μ 2 p ⋅ε μ 4 −=ipppmfi 2 ()13 + 2 qpk2 4⋅ 3 (3) +2ie μ −=ippm () +με fi q2 13 Note, that the sum of all three terms is “gauge invariant”, i.e. it vanishes when ε//k: For instance, take ε=αk, with α some constant, then: 3 ε =⇒++=ak(1) (2) (3)2ie ( p + p ) μ ⎡⎤ apμμμ −+ ap ak q2 13⎣⎦ 4 2 33μμ = 22iiie ap()+−+=+−= p⎡⎤ pμ pμμ μ kie ap ()()0 p p p qq2213⎣⎦ 4 2 13 13 because of energy‐momentum conservation. Note that the sum is zero when ε//k, but not the individual terms sepyparately. The gggauge invariance principle applies to the full amplitude and so includes all orders in e. However, because e is an –a priori unknown – parameter, it is reasonable to assume that gauge invariance must hold order‐by‐order in e. 10/19/2009 44